Why does $\int f(x) dx = \int_0^x f(t) dt$?

Question

I feel like I am missing something fundamental as to why $$\int f(x) dx = \int_0^x f(t) dt$$

Sorry if this is a basic question. I just really want to know why they are equal and why we even have indefinite integrals if they are the same.

Answer 1

$\int f(x) dx$ isn't one function, but a collection of functions. $\int_0^x f(t)dt$ is one of those functions, as is $\int_3^x f(t) dt$. Some times, there are even functions which are not representable as $\int_a^x f(t) dt$.

Example: Take $f(x)=x$. Then $\int f(x) dx$ is the collection of functions of the form $\frac12x^2+C$ for all possible real numbers $C$. On the other hand, $\int_0^xf(t)dt=\frac12x^2$ is one such function, as is $\int_3^xf(t)dt=\frac12x^2-4.5$.

The function $\frac12x^2+1$ is also one of the functions included in $\int f(x)dx$, although it's impossible to construct it as $\int_a^xf(t)dt$ for any real number $a$.

WolframAlpha isn't always good at picking up such subtleties, and should therefore be trusted only to a certain degree with these things.